First, note that $a^{2m} \equiv 1 \pmod{a^m + 1}$, hence if $(m,n)$ is a solution then $(m, n - 2m)$ is a solution given $n \ge 2m$. Hence we can WLOG assume $n < 2m$. We have $(a^m + 1)|(a^n + 203) \implies (a^m + 1)|(a^n - 203a^m) \implies (a^m + 1)|(a^{n-m} - 203)$ due to $\gcd(a^m + 1, a) = 1$. (Note that except for a few exceptional counterexamples (i.e. $a^m < 203$ and possibly some more conditions) we have $n > m$, if I were in this exam I would find them but I'm lazy right now ) However, as we took $n \le 2m$ we have then $a^{n-m} - 203 < a^m + 1$, a contradiction! Hence it suffices to check those counterexamples above. If I find a lot of spare time and remember this thread I will add a detailed report on which work.

We can first think about that n=m+1or2,then we have a^n>203a^m